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यदि $\omega $ इकाई का एक घनमूल हो, तो $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right| = $
${x^3} + 1$
${x^3} + \omega $
${x^3} + {\omega ^2}$
${x^3}$
Solution
(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right|$
=$\left| {\,\begin{array}{*{20}{c}}{x + 1 + \omega + {\omega ^2}}&\omega &{{\omega ^2}}\\{x + 1 + \omega + {\omega ^2}}&{x + {\omega ^2}}&1\\{x + 1 + \omega + {\omega ^2}}&1&{x + \omega }\end{array}\,} \right|$,$({C_1} \to {C_1} + {C_2} + {C_3})$
= $x\,\left| {\,\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\1&{x + {\omega ^2}}&1\\1&1&{x + \omega }\end{array}\,} \right|$ $ (1 + {\omega} + {\omega^2} = 0) $
= $x\,[1\{ (x + {\omega ^2})\,(x + \omega ) – 1\} + \omega \{ 1 – (x + \omega )\} $
$ + {\omega ^2}\{ 1 – (x + {\omega ^2})\} ]$
= $x({x^2} + \omega x + {\omega ^2}x + {\omega ^3} – 1 + \omega – \omega x – {\omega ^2}$
$ + {\omega ^2} – {\omega ^2}x – {\omega ^4})$
= ${x^3}$ , .$(\because \,\,{{\omega }^{3}}=1)$
